3.3.28 \(\int (a+a \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx\) [228]

Optimal. Leaf size=83 \[ \frac {5}{8} a c^3 x+\frac {5 a c^3 \cos ^3(e+f x)}{12 f}+\frac {5 a c^3 \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{4 f} \]

[Out]

5/8*a*c^3*x+5/12*a*c^3*cos(f*x+e)^3/f+5/8*a*c^3*cos(f*x+e)*sin(f*x+e)/f+1/4*a*cos(f*x+e)^3*(c^3-c^3*sin(f*x+e)
)/f

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Rubi [A]
time = 0.08, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2815, 2757, 2748, 2715, 8} \begin {gather*} \frac {5 a c^3 \cos ^3(e+f x)}{12 f}+\frac {a \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{4 f}+\frac {5 a c^3 \sin (e+f x) \cos (e+f x)}{8 f}+\frac {5}{8} a c^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^3,x]

[Out]

(5*a*c^3*x)/8 + (5*a*c^3*Cos[e + f*x]^3)/(12*f) + (5*a*c^3*Cos[e + f*x]*Sin[e + f*x])/(8*f) + (a*Cos[e + f*x]^
3*(c^3 - c^3*Sin[e + f*x]))/(4*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2757

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-b)*(
g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Dist[a*((2*m + p - 1)/(m + p)), Int
[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2,
0] && GtQ[m, 0] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2815

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rubi steps

\begin {align*} \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx &=(a c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^2 \, dx\\ &=\frac {a \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{4 f}+\frac {1}{4} \left (5 a c^2\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x)) \, dx\\ &=\frac {5 a c^3 \cos ^3(e+f x)}{12 f}+\frac {a \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{4 f}+\frac {1}{4} \left (5 a c^3\right ) \int \cos ^2(e+f x) \, dx\\ &=\frac {5 a c^3 \cos ^3(e+f x)}{12 f}+\frac {5 a c^3 \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{4 f}+\frac {1}{8} \left (5 a c^3\right ) \int 1 \, dx\\ &=\frac {5}{8} a c^3 x+\frac {5 a c^3 \cos ^3(e+f x)}{12 f}+\frac {5 a c^3 \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{4 f}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 54, normalized size = 0.65 \begin {gather*} \frac {a c^3 (60 f x+48 \cos (e+f x)+16 \cos (3 (e+f x))+24 \sin (2 (e+f x))-3 \sin (4 (e+f x)))}{96 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^3,x]

[Out]

(a*c^3*(60*f*x + 48*Cos[e + f*x] + 16*Cos[3*(e + f*x)] + 24*Sin[2*(e + f*x)] - 3*Sin[4*(e + f*x)]))/(96*f)

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Maple [A]
time = 0.24, size = 89, normalized size = 1.07

method	result	size
risch	\(\frac {5 a \,c^{3} x}{8}+\frac {a \,c^{3} \cos \left (f x +e \right )}{2 f}-\frac {a \,c^{3} \sin \left (4 f x +4 e \right )}{32 f}+\frac {a \,c^{3} \cos \left (3 f x +3 e \right )}{6 f}+\frac {a \,c^{3} \sin \left (2 f x +2 e \right )}{4 f}\)	\(78\)
derivativedivides	\(\frac {-a \,c^{3} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {2 a \,c^{3} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+2 a \,c^{3} \cos \left (f x +e \right )+a \,c^{3} \left (f x +e \right )}{f}\)	\(89\)
default	\(\frac {-a \,c^{3} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {2 a \,c^{3} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+2 a \,c^{3} \cos \left (f x +e \right )+a \,c^{3} \left (f x +e \right )}{f}\)	\(89\)
norman	\(\frac {\frac {4 a \,c^{3}}{3 f}+\frac {5 a \,c^{3} x}{8}+\frac {4 a \,c^{3} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {4 a \,c^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}+\frac {4 a \,c^{3} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {3 a \,c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}+\frac {11 a \,c^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}-\frac {11 a \,c^{3} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}-\frac {3 a \,c^{3} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}+\frac {5 a \,c^{3} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {15 a \,c^{3} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {5 a \,c^{3} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {5 a \,c^{3} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{4}}\)	\(244\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))*(c-c*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

1/f*(-a*c^3*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-2/3*a*c^3*(2+sin(f*x+e)^2)*cos(f*x+e
)+2*a*c^3*cos(f*x+e)+a*c^3*(f*x+e))

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Maxima [A]
time = 0.31, size = 93, normalized size = 1.12 \begin {gather*} \frac {64 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a c^{3} - 3 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a c^{3} + 96 \, {\left (f x + e\right )} a c^{3} + 192 \, a c^{3} \cos \left (f x + e\right )}{96 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c-c*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

1/96*(64*(cos(f*x + e)^3 - 3*cos(f*x + e))*a*c^3 - 3*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a
*c^3 + 96*(f*x + e)*a*c^3 + 192*a*c^3*cos(f*x + e))/f

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Fricas [A]
time = 0.33, size = 67, normalized size = 0.81 \begin {gather*} \frac {16 \, a c^{3} \cos \left (f x + e\right )^{3} + 15 \, a c^{3} f x - 3 \, {\left (2 \, a c^{3} \cos \left (f x + e\right )^{3} - 5 \, a c^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c-c*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

1/24*(16*a*c^3*cos(f*x + e)^3 + 15*a*c^3*f*x - 3*(2*a*c^3*cos(f*x + e)^3 - 5*a*c^3*cos(f*x + e))*sin(f*x + e))
/f

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (78) = 156\).
time = 0.21, size = 196, normalized size = 2.36 \begin {gather*} \begin {cases} - \frac {3 a c^{3} x \sin ^{4}{\left (e + f x \right )}}{8} - \frac {3 a c^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - \frac {3 a c^{3} x \cos ^{4}{\left (e + f x \right )}}{8} + a c^{3} x + \frac {5 a c^{3} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {2 a c^{3} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {3 a c^{3} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {4 a c^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac {2 a c^{3} \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (a \sin {\left (e \right )} + a\right ) \left (- c \sin {\left (e \right )} + c\right )^{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c-c*sin(f*x+e))**3,x)

[Out]

Piecewise((-3*a*c**3*x*sin(e + f*x)**4/8 - 3*a*c**3*x*sin(e + f*x)**2*cos(e + f*x)**2/4 - 3*a*c**3*x*cos(e + f
*x)**4/8 + a*c**3*x + 5*a*c**3*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 2*a*c**3*sin(e + f*x)**2*cos(e + f*x)/f +
3*a*c**3*sin(e + f*x)*cos(e + f*x)**3/(8*f) - 4*a*c**3*cos(e + f*x)**3/(3*f) + 2*a*c**3*cos(e + f*x)/f, Ne(f,
0)), (x*(a*sin(e) + a)*(-c*sin(e) + c)**3, True))

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Giac [A]
time = 0.44, size = 81, normalized size = 0.98 \begin {gather*} \frac {5}{8} \, a c^{3} x + \frac {a c^{3} \cos \left (3 \, f x + 3 \, e\right )}{6 \, f} + \frac {a c^{3} \cos \left (f x + e\right )}{2 \, f} - \frac {a c^{3} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {a c^{3} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c-c*sin(f*x+e))^3,x, algorithm="giac")

[Out]

5/8*a*c^3*x + 1/6*a*c^3*cos(3*f*x + 3*e)/f + 1/2*a*c^3*cos(f*x + e)/f - 1/32*a*c^3*sin(4*f*x + 4*e)/f + 1/4*a*
c^3*sin(2*f*x + 2*e)/f

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Mupad [B]
time = 9.00, size = 250, normalized size = 3.01 \begin {gather*} \frac {5\,a\,c^3\,x}{8}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a\,c^3\,\left (15\,e+15\,f\,x\right )}{6}-\frac {a\,c^3\,\left (60\,e+60\,f\,x+32\right )}{24}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {a\,c^3\,\left (15\,e+15\,f\,x\right )}{6}-\frac {a\,c^3\,\left (60\,e+60\,f\,x+96\right )}{24}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {a\,c^3\,\left (15\,e+15\,f\,x\right )}{4}-\frac {a\,c^3\,\left (90\,e+90\,f\,x+96\right )}{24}\right )-\frac {3\,a\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}-\frac {11\,a\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{4}+\frac {11\,a\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{4}+\frac {3\,a\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{4}+\frac {a\,c^3\,\left (15\,e+15\,f\,x\right )}{24}-\frac {a\,c^3\,\left (15\,e+15\,f\,x+32\right )}{24}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))*(c - c*sin(e + f*x))^3,x)

[Out]

(5*a*c^3*x)/8 - (tan(e/2 + (f*x)/2)^2*((a*c^3*(15*e + 15*f*x))/6 - (a*c^3*(60*e + 60*f*x + 32))/24) + tan(e/2
+ (f*x)/2)^6*((a*c^3*(15*e + 15*f*x))/6 - (a*c^3*(60*e + 60*f*x + 96))/24) + tan(e/2 + (f*x)/2)^4*((a*c^3*(15*
e + 15*f*x))/4 - (a*c^3*(90*e + 90*f*x + 96))/24) - (3*a*c^3*tan(e/2 + (f*x)/2))/4 - (11*a*c^3*tan(e/2 + (f*x)
/2)^3)/4 + (11*a*c^3*tan(e/2 + (f*x)/2)^5)/4 + (3*a*c^3*tan(e/2 + (f*x)/2)^7)/4 + (a*c^3*(15*e + 15*f*x))/24 -
(a*c^3*(15*e + 15*f*x + 32))/24)/(f*(tan(e/2 + (f*x)/2)^2 + 1)^4)

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